Question

llusA wrecking ball on a 15.4 m longcable is pulled back to an angle of33.5° and released. At what speedis it moving at the bottom of itsswing?(Unit = m/s)

Answer 1

In order to calculate the speed of the ball at the bottom, first, let's calculate it's initial height, using the cosine of the angle:

$\begin{gathered} \cos (33.5)=\frac{15.4-h}{15.4} \\ 0.834=\frac{15.4-h}{15.4} \\ 15.4-h=12.844 \\ h=2.556 \end{gathered}$

Now, let's convert the initial gravitational energy into kinetic energy:

$\begin{gathered} m\cdot g\cdot h=\frac{m\cdot v^2}{2} \\ 9.81\cdot2.556\cdot2=v^2 \\ v^2=50.149 \\ v=7.08 \end{gathered}$

So the speed at the bottom is 7.08 m/s.